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Chapter 7

Q. 7.5

A rectangular beam 300 mm deep is simply supported over a span of 4 metres. Determine the uniformly distributed load per metre which the beam may carry, if the bending stress should not exceed 120 N/mm². Take I = 8 × 10^6  mm^4.

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Verified Solution

Given :
Depth,       d = 300 mm
Span,         L = 4 m
Max. bending stress, σ_{\max} = 120  N/mm^2
Moment of inertia,    I = 8 × 10^6  mm^4
Let, w = U.D.L. per metre length over the beam in N/m.

The bending stress will be maximum, where bending moment is maximum. For a simply supported beam carrying U.D.L., the bending moment is maximum at the centre of the beam [i.e., at point C of Fig. 7.11 (b)]

∴    Max. B.M. = 2w × 2 – 2w × 1 = 4w – 2w
= 2w Nm \left(\text { Also } M=\frac{w \times L^2}{8}=\frac{w \times 4^2}{8}=\frac{16 w}{8}=2 w\right)
= 2w × 1000 Nmm
or           M = 2000w Nmm

Now using equation (7.6), we get

M=\sigma_{\max } \times Z             …(i)

where          Z=\frac{I}{y_{\max }}=\frac{8 \times 10^6}{150}\quad  \left(\because  y_{\max }=\frac{d}{2}=\frac{300}{2}=150  mm \right)

Hence above equation (i) becomes as

2000 w=120 \times \frac{8 \times 10^6}{150}

or            w=\frac{120 \times 8 \times 10^6}{2000 \times 150}=\pmb{3200  N / m.}

7.11b